Sextupole

Kai Hock

Cockcroft Institute / Liverpool University

June 22, 2010

The sextupole magnet is needed to correct for the chromatic effect of a quadrupole. Consider a quadrupole that focuses a beam of in the x direction to a focal point. If some of the electrons have higher energies, they would be bent less by the quadrupole field, and their focal points would be further away, as illustrated in fig. 1. The result of this is that a bunch of electrons would get spread out or defocused in the longitudinal direction.

In order to correct for this chromatic effect, a sextupole magnet may be used. The job of a sextupole is illustrated in fig. 2. Compared with a quadrupole, a sextupole must have a larger focusing effect for particles that are displaced further from the axis. This happens with the sextupole because it is designed to have a magnetic field that varies as x2, rather than the x1 in the quadrupole.

1 Figure 1: Chromatic effect.

Figure 2: How a sextupole magnet works.

2 Figure 3: A sextupole magnet in Australian [1].

A photo of a real sextupole is shown in fig. 3. The field is shown schemat- ically in fig. 4.

Here, I shall go into how the sextupole works in detail, but only explain how the sextupole field leads to the elements in the transfer matrix. The sextupole field is a bit more complex than the dipole and quadrupole fields.

So it is convenient that there is one formula that can be used to derive all these fields. This is the multipole expansion.

We start with the magnetic scalar potential ψ, given by [2]:

B = −µ0∇ψ (1)

where B is the magnetic field. For convenience, µ0ψ will be replaced with φ.

A multipole field expansion for the scalar field can be written as [6]:

∞ X m φ = r {Am cos mθ + Bm sin mθ} (2) m=0

3 Figure 4: The field in a sectupole magnet [3]. where r is the distance from the axis, θ the azimuthal angle, and θ = 0 is the positive x direction. Am and Bm are constant coefficients. Lets look at the magnetic field arising from each term.

When m = 0, it is a constant term:

φ = A0 (3) so the field is zero:

B = −∇φ = 0i + 0j (4)

When m = 1,

φ = A1r cos θ + B1r sin θ (5)

In Cartesian co-ordinates, it is:

φ = A1x + B1y (6)

4 So we get the dipole field:

B = −∇φ = −A1i − B1j (7)

When m = 2:

2 2 φ = A2r cos 2θ + B2r sin 2θ (8) which is the same as:

2 2 φ = A2(x − y ) + B22xy (9)

This gives the quadrupole field:

B = −∇φ = A2(−2xi + 2yj) + B2(−2yi − 2xj) (10)

A2 = 0 gives the normal quadrupole, and B2 = 0 gives the skew quadrupole.

When m = 3:

3 3 φ = A3r cos 3θ + B3r sin 3θ (11)

Rewriting this in Cartesian co-ordinates, the sextupole can be derived:

2 2 2 2 B = −∇φ = A3(−3(x − y )i + 6xyj) + B3(−6xyi − 3(x − y )j) (12)

For the normal sextupole, A3 = 0, and the x and y magnetic field com- ponents are:

Bx = −6B3xy (13)

2 2 By = −3B3(x − y ) (14)

5 The transfer matrix elements in [4] is written in terms of the second derivative of the field, so I shall do this now. Differentiating:

∂2B y = −6B (15) ∂x2 3

In terms of this:

∂2B B = y xy (16) x ∂x2 1 ∂2B B = y (x2 − y2) (17) y 2 ∂x2

We are now ready to derive the elements for the sextupole. The sextupole is simply treated as a kicker. Thus, By produces a horizontal force and is modelled as a horizontal kicker. Likewise, the effect from Bx is modelled as a vertical kicker.

Then the elements for a kicker can be used directly, and the expressions for the sextupole field substituted.

The horizontal kicker elements are [5]:

 l  m = −m = −m = B (18) 26 51 27 Bρ y  l  2 1  m = −C E3 B (19) 67 γ 0 Bρ y 2πl 1 l   m = B m (20) 27 2 Bρ y 67

Substituting the expressions for By in eq. (17), we get:

 l 1 ∂2B m = −m = −m = y (x2 − y2) (21) 26 51 27 Bρ 2 ∂x2

6  l 1 ∂2B 2 1  m = −C E3 y (x2 − y2) (22) 67 γ 0 Bρ 2 ∂x2 2πl 1 l 1 ∂2B  m = y (x2 − y2) m (23) 27 2 Bρ 2 ∂x2 67

Using the symbol defined in [4]:

l ∂2B λ = y (24) Bρ ∂x2 we obtain the required sextupole transfer matrix elements:

1 m = −m = −m = λ(x2 − y2) (25) 26 51 27 2 1 2 1  m = −C E3 λ(x2 − y2) (26) 67 γ 0 2 2πl 11  m = λ(x2 − y2) m (27) 27 2 2 67

The element m27 is not given in [4], possibly because it is small.

The values of x and y just before the sextupole are x0 and y0, so:

1 m = −m = −m = λ(x2 − y2) (28) 26 51 27 2 0 0 1 2 1  m = −C E3 λ(x2 − y2) (29) 67 γ 0 2 0 0 2πl

So far, we have derived the matrix elements for the effects of By using the formulae for the horizontal kicker. The matrix elements for the effects of Bx can be derived similarly, using the formulae for the vertical kicker [4], and the expression for Bx in eq. (17). The first few elements, for example, are given by:

−m46 = m53 = m47 = λx0y0 (30)

7  2 1  m = −C E3 λx y (31) 67 γ 0 0 0 2πl

Note that there is a common term m67 in eqs. (29) and (31). They are separate contributions from By and Bx respectively. So the combined effect would be the sum. Adding, we get:

 1  m = −C E3λ2(x2 + y2)2 (32) 67 γ 0 0 0 8πl

References

[1] http://en.wikipedia.org/wiki/Sextupole_magnet

[2] http://en.wikipedia.org/wiki/Magnetic_potential

[3] Andy Wolski’s linear dynamics lecture 3, http:http://hep.ph.liv.

ac.uk/~hock/Good_References/Good_References.html

[4] Alex Chao, Evaluation of Beam Distribution Parameters in an Elec-

tron , SLAC-PUB-2143. http://www.slac.stanford.

edu/pubs/slacpubs/2000/slac-pub-2143.html

[5] http://hep.ph.liv.ac.uk/~hock/Damping_Ring/kicker_magnet.

html

[6] P. J. Bryant and K. Johnsen, Circular Accelerators and Storage Rings,

(Cambridge University Press, UK) 1993, pp. 317 - 319.

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